1. ## Integration by principles

I need to calculate the following integral by principle, that means resolving the limit of the sum.

The integration segment [0,T] is partitioned in N partitions of size T/N.

$
\int_0^{T} x^4\ dx\ = \lim_{N\to \infty}\sum_{i=0}^{N-1} x_{i}^{4}(x_{i+1} - x_{i})
$

Thanks a bunch!

2. I suppose you're referring to a Riemann sum.

${\Delta}x=\frac{T}{n}$

Using the right endpoint method, $a+k{\Delta}x$

Since a=0, we have $\frac{kT}{n}$

Sub into what is to be integrated:

$\left(\frac{kT}{n}\right)^{4}\cdot\frac{T}{n}$

$\frac{t^{5}}{n^{5}}\sum_{k=1}^{n}k^{4}$

The thing to do here is to write the sum in terms of n.

The sum of the first n fourth powers is given by:

$\frac{n^{5}}{5}+\frac{n^{4}}{2}+\frac{n^{3}}{3}-\frac{n}{30}$

So, you get:

$\frac{t^{5}}{n^{5}}\left(\frac{n^{5}}{5}+\frac{n^{ 4}}{2}+\frac{n^{3}}{3}-\frac{n}{30}\right)$

Expand out and take the limit as $n\rightarrow{\infty}$

You should get your answer the same as if you integrated the 'quick' way.

3. ## Could be done easier?

There must be some trick...

For example:

$
\int_0^{T} 2x\ dx\ = \lim_{N\to \infty}\sum_{i=0}^{N-1} 2x_{i}(x_{i+1} - x_{i}) = x^2(T) - T
$

That was done using the trick 2ab = [(a+b)^2 - a^2 - b^2] with a=x(i) and b=[x(i+1) - x(i)]

The series becomes telescopic and yields the results...